Ink on Sequences #6

Similar solution with @H.M. 유 's

$\begin{aligned} a_{n+1} & = \frac {a_n^2 - a_{n-1}+2a_n}{a_{n-1}+1} \\ & = \frac {a_n^2+2a_n {\color{#3D99F6}+ 1}-a_{n-1}\color{#3D99F6}-1}{a_{n-1}+1} \\ & = \frac {(a_n+1)^2 -(a_{n-1}+1)}{a_{n-1}+1} \\ & = \frac {(a_n+1)^2}{a_{n-1}+1} - 1 \\ \implies a_{n+1} + 1 & = \frac {(a_n+1)^2}{a_{n-1}+1} & \small \color{#3D99F6} \text{Let }b_k = a_k + 1 \\ b_{n+1} & = \frac {b_n^2}{b_{n-1}} \\ \implies b_{n-1}b_{n+1} & = b_n^2 \end{aligned}$

This means that $\{b_n\}$ is a geometric progression and therefore,

$\begin{aligned} b_{2017} & = b_1r^{2017-1} & \small \color{#3D99F6} \text{where }r \text{ is the common ratio.} \\ a_{2017} + 1 & = (a_1 + 1)r^{2016} \\ 2018 & = 2r^{2016} \\ \implies r & = \sqrt [2016]{1009} \\ b_2 & = b_1r = 2 \sqrt[2016]{1009} \\ \implies a_2 & = -1 + 2 \sqrt[2016]{1009} \end{aligned}$

$\implies \alpha + \beta + \gamma + \delta + \epsilon = -1+0+2+2016+1009 = \boxed{3026}$

Let $b_n=a_n+1.$

Then,

$\begin{aligned} a_{n+1} &=\dfrac{(a_n+1)^2-a_{n-1}-1}{a_{n-1}+1} \\ b_{n+1}-1 &=\dfrac{{b_n}^2-b_{n-1}}{b_{n-1}} \\ b_{n+1} &= \dfrac{{b_n}^2}{b_{n-1}} \\ \therefore~b_{n-1}b_{n+1} &={b_n}^2 \end{aligned}$

This shows that $b_n$ is a geometric progression .

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$b_1=2$ and $b_{2017}=2018,$ and therefore if we let the common ratio be $r,$

$r^{2016}=1009$

$\therefore~r=\sqrt[2016]{1009}.$

Then $b_2=2\cdot\sqrt[2016]{1009}.$

Therefore, $a_2=b_2-1=-1+2\cdot\sqrt[2016]{1009}.$

$\alpha+\beta+\gamma+\delta+\epsilon=-1+0+2+2016+1009=\boxed{3026}.$